GAMSAT 1 (High Value Chem Topics)

Question 1

*

Using k_w to find the pH of a Base

Answer 1

k_w=[H⁺][OH]

NaOH & KOH - Strong bases - donate 1 mole of OH^- ions per mole of base

• the value of kw @ 298K us 1.0 X 10^-14 mol²dm^-6
• find the pH of 0.1 moldm^-3 NaOH @298K
  
  • [OH^-] = 0.1moldm^-3 => [H⁺] = k_w/[OH^-] = 1.0 X 10^-14/.01
  • pH= -log10 1.0 X 10^-13 = 13

Question 2

*

Diprotic Acids

Answer 2

They release 2 protons when it dissociates

ex. H₂SO_4(l)+water -> 2H⁺_(aq) + SO₄^-2_(aq)

H⁺ = 0.2 moldm^-3 so….pH = -log₁₀[0.2] = 0.70

pH of sulfuric acid 0.25 moldm^-3

[H⁺] = 2 X 0.25 = 0.5 => pH = ^-log₁₀[.5] = 0.30

Question 3

*

pH Definition

Answer 3

• The measure (from 0-14) of Hydrogen Ion Concentration
  
  • 0 = Very Acidic
  • 7 = Neutral
  • 14 = Very Alkaline (base/basic)

*Expressed in -log₁₀ —> pH = -log₁₀[H⁺] ex. pH = -log₁₀[0.01] = 2

Or…

[H⁺] = 10^-pH ex. [H⁺] = 10^-1.52 = 0.03moldm^-3 = 3 X 10^-2moldm^-3

Question 4

*

Acids & Bases

Answer 4

• Protons are transferred (Acid -> Bases) when A&B react
• Acids can olny get rid of protons when there is a base to accept them
  
  • ex. HA_(aq)+B_(aq) ⇔BH⁺_(aq)+ A^-_(aq)
• If acid is addded to water the water acts as the base and accepts the proton
  
  • HA_(aq) + H₂O ⇔ H₃O⁺_(aq) + A^-_(aq)

Question 5

*

Bases: Strong & Weak

Answer 5

• Strong Bases - ionise almost completely in water too
  
  • ex. sodium hydroxide - NaOH_(s) + Water ⇒ Na⁺_(aq) + OH^-_(aq)

_​

• Weak Bases - only slightly dissociate in water
  
  • ex. Ammonia - NH_3(aq) + H₂0 ⇔ NH₄⁺_(aq) + OH^-_(aq)

_{​*Equilibrium lies well over to the left}

Question 6

*

Acids: Strong & Weak

Answer 6

• Acid releases a proton - A base accepts a proton

Strong Acids - dissociate (or ionise) almost completely in water - nearly all the H⁺ ions will be released

(Hydrochloric acid) ex. HCl_(g) + Water ⇒ H⁺_(aq) + Cl^-_(aq)

Weak Acids - dissociate only very slightly in water - so only small numbers of H⁺ ions are formed

(Ethanoic or citric) ex. CH₃COOH_(aq) ⇔CH₃COO^-_(aq) + H⁺_(aq)

Question 7

*

Kinetics: Rate Equation

Answer 7

Rate = k [A]ⁿ + [B]^m

^[A] - Concentration of A

n - rate order

k - Rate Constant

*Overall order of reaction is n + m

* Increase in temperature will increase k

Question 8

*

Orders of Reaction

Answer 8

* Order can be determined only by experiment using the Method of Initial Rates (the rate for a short time @ the beginning of the reaction is measured @ several different concentrations of reactants)

1. First Order - X¹ = Rate doubles when reactant doubles
   
   • X2 = X2; X3 = X3; …
2. Second Order - X² = Rate is (x4) when the reactant doubles
   
   • X2 = X4 (2²) ; X4 = X16; …..
3. Zero Order - X⁰ = rate stays the same regardless of reactant
   
   • X2 = 1; X4 = 1; …..

Question 9

*

Kinetics: Concentration of Catalysts

Answer 9

• Increase concentration of catalysts = Increase rate of reaction
• Increase concentration of reactants in a solution, on avg. the particles will be closer together
  
  • closer ⇒ collide more often ⇒ more collisions ⇒ more chances to react
• if gases are involved, and increase in pressure of the gas works the same way
• Catalysts increase rate of reactions too by providing an alternative reaction pathway w/ a lower activation energy​

*Catalyst is chemically unchanged @ the end

Question 10

*

Kinetics: Physical State of Reactants

Answer 10

• Particles must collide to react
  
  • in the right direction, facing the right way
  • must collide w/ the min. amt. of kinetic energy

(Collision Theory)

*Liquids & Gases best as particles move

*Increase in temp. = particles have more kinetic energy = faster reactions @ activation energy more particles have enough energy @ 35ºC > 25ºC

Question 11

*

Kinetics: Physical State of Reactants

Answer 11

• Solids - particles very close together.
  
  • High density & incompressible.
  • Particles vibrate about a fixed point & can’t move freely
• Liquids - Similar density to a solid & is virtually incompressible
  
  • Particles move freely & randomly w/in the liquid
• Gas - particles have lots more energy & are much further apart
  
  • Density is pretty low & it’s very compressible
  • Particles move freely, diffuse quickly, no alot of attraction between them

Question 12

*

Gibbs free energy (1)

Answer 12

• Free enthalpy, Gibbs energy, or Gibbs function
• ΔG = ΔH - TΔS
  
  • H - Heat energy in the system (kj)
  • S - Measure of Entropy (J/kmol)
  • T - temp (K)​​
    
    • Equation to determine how likely a reaction is to take place spontaneously

* If a reaction will take place it reduces Gibbs free energy (ΔG < 0)

*Gibbs energy is reduced if H is reduced

*Gibbs energy is reduced if S is increased

Question 13

*

Gibbs free energy (2)

Answer 13

• Reactions most likely to happen if: ΔH < 0 & ΔS > 0

Endo & Exothermic

• Endo - energy taken in so heat in system is increased ΔH > 0 = unfavourable
• Exo - energy given off so heat in system is reduced ΔH < 0 = favourable

* Endothermic reaction can still take place if it results in a large enough increase in entropy (ΔS>0)

Question 14

*

Gibbs free energy (3)

Answer 14

* at a phase change (gas → water )

• ΔG = 0
• ΔG = positive # = not spontaneous
• ΔG = negative # = spontaneous

ex. ΔG = ΔH -TΔS

• • • : ΔG= negative if temp high
• • • : ΔG= negative if temp low
• • • : ΔG= positive (non spontaneous) always
• • • : ΔG= negative always

Question 15

*

Feasibility of Reactions

Answer 15

• More Negative ( or less positive) E^{<span>Ø</span>} value moves left
• More Positive (or less negative) E^{<span>Ø</span>} value moves right

Ex. Fe(OH)_3(s)+ e^- ⇔ Fe(OH)_2(s) + OH^-_(aq) E^{<span>Ø</span>} = -0.56V

O_2(g) + 2H₂O + 4e^- ⇔ 4OH^-_(aq) E^Ø = ⁺0.40V

Question 16

*

Electrode Potential

(conditions affecting it)

Answer 16

• Half cell reactions are reversible
  
  • equilibrium position is affected by changes in:
    
    1. Temperature
    2. Pressure
    3. Concentration
• Standard Conditions are:
  
  1. Temp - 25°C (298K)
  2. Pressure - 100kPa
  3. Concentration - 1.00 moldm^-3

Question 17

*

Electrode Potential

(standard elctrochemical cell drawings)

Answer 17

• the potential difference between the electrode & its solution

E^øcell = (E^ø_{right side} - E^ø_{left side})

Zn/Cu cell short hand

Zn_(s) | Zn²⁺_(aq) || Cu²⁺_(aq) | Cu_(s) (Zn²⁺_(aq) + 2e^- ⇔ Zn_(s) )

—Charges go this way——► (Cu²⁺_(aq) + 2e^- ⇔ Cu_(s))

Reduced| Oxidised || Oxi. | Red.

Question 18

Substitution & Elimination (1)

(Rules & what influences then)

Answer 18

Most important factor: type of Halokane

• Primary - mostly substitution
• Secondary - both substitution & elimination
• Tertiary - mostly elimination
  
  • Can be influenced by changing conditions​

The Solvent = proportion of ethanol to water

• more water = more substitution
• more ethanol = more elimination

​Concentration - of sodium or potassium hydroxide solution

• higher concentration = higher elimination

Question 19

Substitution & Elimination

Reactions

Answer 19

Substitution - the halogen is replaced by an -OH group to give alcohol

• CH₃CHCH₃ ⇒ NaOH ⇒ CH₃CHCH₃ + NaBr
  
  • | |
  • Br OH

Elimination - also in the presence of Sodium &/or potassium

• hydrogen bond is removed from one of the end carbon atoms toghether w/ brownie from centre one
• ​CH₃CHCH₃ + NaOH ⇒ CH₂=ChCH₃ + NaBr + H₂O
  • ​|
  • Br

Question 20

S_N1 Reactions

Answer 20

S_N1 - Nucleophillic substitution

• S = Substitution ; N = Nucleophilic ; 1 = the initial stage involves 1 species
• Faster mechanism
• best with tertiary halokanes
  
  • ​ex. R₃C - X ⇒ R₃C + ⇒ R₃C - Nu
    
    • ​ ↑
    • Nu

Question 21

S_N2 Reactions

Answer 21

S_N2 - best with Primary Halokanes (initial stage 2 species)

• _X
• ex. R - X → R< → R - Nu
• ^Nu

Question 22

Fischer Projections

Answer 22

• 2-D representation of a 3-D organic molecule
• Used for carbohydrates but not non-carbohydrates ex. Fischer
• Rules H O
  
  • Carbon Chain - vertical, C1 at top \ //
  • Horizontal bonds project toward the viewer C
  • vertical bonds project away |
  • H — C — OH
  • |
  • OH — C — H
  • |
  • H — C — OH
  • |
  • H

Question 23

Optically Active

Answer 23

• Rotate plane - polarised light
  
  • normal light vibrates in all directions but plane - polarised light only vibrates in one direction
• One enantiomer rotates it in a clockwise direction & the other rotates it in an anti clockwise direction

Question 24

Racemates (Isomers)

Answer 24

• Racemic mixture
• Contains equal quantities of each enatiomer of an optically active compound
• don’t show any optical activity
  
  • ​they cancel each other’s light-rotating effect
• ​made typically by reactive 2 achrial things together to get a racemic mixture as the chances of each enantiomer is equal
  
  • ​ex. H CL H
  • | | |
  • C + Cl₂ → HCl + C or C
  • / | \ / | \ / | \
  • CH₃ | H CH₃ | H CH₃| Cl
  • C2H3 C2H3 C₂H₃

Question 25

Chirals

Answer 25

• Carbon w/ 4 different groups attached
• ex. H H O
• | | /
• H—C—C*—C * = Chiral Centre
• | | \
• H OH OH

​Enantiomers:

• H | H
• | | |
• C | C
• / | \ | / | \
• HOOC | CH₃ | H₃C | COOH
• OH | OH

Question 26

Optical Isomers

Answer 26

- type of stereoisomerism - same structural formula but atoms are arranged differently

-Chiral - (asymmetric) carbon atom is an optical isomer that has 4 different groups attached to it. The groups attached to it. The groups can be arranged in 2 different ways so that 2 different molecules are made

* called enantiomers or optical isomers

• ex. H | H
• | | |
• C | C
• / | \ | / | \
• HOOC | CH₃ | H₃C | COOH
• OH | OH

Question 27

Stereoisomers

Answer 27

• Same structural formula but a different arrangement in space
• w/ a double bond molecules can’t twist around so you get either E or Z isomers
• E (entgegen(opposite)) - H CH₃
• \ /
• C=C E-but-2-ene
• / \
• H₃C H
• Z (zusammen (together)) - H₃C CH₃
• \ /
• C=C Z-but-2-ene
• / \
• H H

Question 28

Isomer Formulas

Answer 28

• General Formula - C_nH_2n+1OH
• Empirical (simplest ratio) - C₄H₁₀O
• Molecular (actual #) - C₄H₁₀O
• Structural- CH₃CH₂CH₂CH₂OH
• H H H H
• | | | |
• Displayed H–C–C–C–C–O–H
• | | | |
• H H H H
• Skeletal - shows the bonds of carbon only, w/ any functional group
• /\/_OH

Question 29

Isomers

Answer 29

• Molecules with the same molecular formula but different molecular structures

3 Types

• Chain Isomers - straight & branched carbon skeletons (
  
  • ex. butane | | | |
  • H– C–C–C–C–H
  • | | | |
• Positional Iso - same skeleton but w/ a functional group attached
  
  • ex. l-Chlorobutane | | | |
  • H–C–C–C–C–Cl
• Functional group Isomers - same atoms arranged into functional groups
  
  • ex. propanone H O H
  • | || |
  • H–C–C–C–H
  • | |

Question 30

Periodic Table:

Atomic Radius & Ionisation

Answer 30

• Atomic radius decreases across a period
• ↑ # of Protons = ↑ positive charge of nucleus = ↑ pull of centre = ↓ of radius
• Ionisation ↑ across a period
• ↑ attraction between outer shell & nucleus

Question 31

Periodic Table: Blocks

Answer 31

• blocks show sub-shells configuration
  
  • 1s |H|
  • 2sΠΠ 2pΠΠΠΠΠΠ
  • 3sΠΠ 3pΠΠΠΠΠΠ
  • 4sΠΠ 3dΠΠΠΠΠΠΠΠΠΠΠ 4pΠΠΠΠΠΠ
  • 5sΠΠ 4dΠΠΠΠΠΠΠΠΠΠΠ 5pΠΠΠΠΠΠ
  • 6sΠΠ 5dΠΠΠΠΠΠΠΠΠΠΠ 6pΠΠΠΠΠΠ
  • 7sΠΠ 6dΠ
  • ↖83-103
    
    • s-block - have an outershell config of s¹ or s²
      
      • ​Lithium (1s²2s¹) & Magnesium (1s²2s²2p⁶3s²)
    • p-block - outhershell of s²p¹ → s²p⁶
      
      • Chlorine (1s²2s²2p⁶3s²3p⁵)
    • d-block - d sub-shells are filled
      
      • Cobalt (1s²2s²2p⁶3s²3p⁶3d⁷4s²)

Question 32

Electron Shells:

Subshells & Orbitals

Answer 32

• Subshells are divided into orbitals
• Orbitals can hold up to 2 electrons
• Subshell | # of Orbitals | Max # of electrons
• s | 1 | 1 X 2 = 2
• p | 3 | 3 X 2 = 6
• d | 5 | 10
• f | 7 | 14
• Shell | Subshells | Total # of electrons
• 1st | 1s | 2
• 2nd | 2s 2p | 2+(3x2) = 8
• 3rd | 3s 3p 3d | 18
• 4th | 4s 4p 4d 4f | 32

Question 33

Periodic Table

Trends: periods, groups & atomic #

Answer 33

• all elements in a period (row) have the same # of electron shells
  
  • (ignore s&p subshells)
  • |H| 0
• G1 2 3 4 5 6 7Π
• 2ΠΠ ΠΠΠΠΠΠ
• 3ΠΠ ΠΠΠΠΠΠ
• 4ΠΠΠΠΠΠΠΠΠΠΠΠΠΠ
• 5ΠΠΠΠΠΠΠΠΠΠΠΠΠΠ
• 6ΠΠΠΠΠΠΠΠΠΠΠΠΠΠ
• 7ΠΠΠ
  
  • All groups have same # of electrons in outer shells
    
    • G3 = 3 electrons in outer shells (G1→7,G0)
    • Similar outer shells = similar properties

Question 34

Buffer (pH)

Answer 34

• A solution that resists changes in pH when small amounts of acid or alkalai are added
  
  • doesn’t stop the pH from changing but it does make it very slight
    
    • only works with small amounts of acids or bases
  • salt is a common buffer as it takes a few of the H⁺ ions

Question 35

Indicators (pH)

Answer 35

• Change colour at range that lies entirely on the vertical part of the pH curve
• Methyl orange & Phenolphthalein are common indicators
• Methyl orange - | Start | 3.1 - 4.4pH | High pH |
• | Red | Orange | Yellow |
• • Phenolphthalein | Start | 8.3 - 1.0pH | High pH |
• | White | White/Pink | Pink |
• • Weak base/ weak acid - no sharp curve → indicators won’t work
• Strong base/ strong acid - Either is fine

Question 36

8 Strong Bases

Answer 36

1. LiOH - Lithium hydroxide
2. NaOH - Sodium hydroxide
3. KOH - Potassium hydroxide
4. RbOH - Rubidium hydroxide
5. CsOH - Cesium hydroxide
6. Ca(OH)₂ - Calcium hydroxide
7. Sr(OH)₂ - Strontium hydroxide
8. Ba(OH)₂ - Barium hydroxide

Question 37

7 Strong Acids

Answer 37

1. HCl - Hydrochloric acid
2. HNO₃ - Nitric acid
3. H₂SO₄ - Sulfuric acid
4. HBr - Hydrobromic acid
5. HI - Hydroiodic acid
6. HClO₃ - Chloric acid
7. HClO₄ - Perchloric acid

Question 38

Naming acids

Answer 38

• Binary acids - 2 elements
  
  • (prefix) hydro + compound + -ic
    
    • ex. HCl = hydrochloric acid
• Tertiary acids - typically hydrogen + a non-metal + oxygen
  
  • named by Oxygen amount
  • 2 common form = (prefix) hypo + compound + (suffix) -ous
    
    • ex. Hypochlorous acid | HClO
  • 1 common form = suffix = -ous
    
    • ex. Chlorous acid | HClO₂
  • most common form = suffix -ic
    
    • ex. Chloric acid | HClO₃
  • +1 common form = (prefix) -per + compound + (suffix) -ic
    
    • ex. Perchloric acid | HClO₄

Question 39

pH Curves

Answer 39

Question 40

Calculate pK_a & also pH from pK_a

Answer 40

• pK_a = -log_{10<span>K</span>a} & k_a = 10^-pK_a
• pH = -log₁₀[H⁺] & [H⁺] = 10^-pH
• • ex. Calculate the pH of 0.050 moldm^-3 HCOOH
    * pK_a of 3.75 @ 298k
    * K_a = 10^-pK_a = 10^-3.75 = 1.78 X 10^-4 moldm^-3
    
    • • K_a = [H⁺]²/[HCOOH] ⇒ [H⁺]² = K_a[HCOOH] = 1.78 X 10^-4 X 0.050 =8.9 X 10^-6
        
        • ^​⇒[H⁺] = √8.9 X 10^-6 = 2.98 X 10^-3 moldm^-3
        • pH = -log₁₀ 2.98 X 10^-3 = 2.53

Question 41

Calculate [H⁺] from pH

Answer 41

• [H⁺] = 10^-pH = ? X 10 ^-? moldm^-3
• • ex. pH = 3.02
    
    • [H⁺] = 10^-3.02 = 9.55 X 10^-4 moldm^-3
      *

Question 42

Dissociation Constant

Answer 42

• used to find the pH of a weak acid
  
  • K_a = _[H⁺][A^-] ex. HA_(aq) ⇔ H⁺_(aq) + A^-_(aq)
  • ^{—————} *can assume that all acids come from the
  • ^[HA] acid so [H⁺] = [A^-] ⇒ [H⁺]²
  • • ​K_a = _[H⁺]²
• ^——— pH = -log₁₀[H⁺]
• ^[HA]

Question 43

Electrochemical Cells

Answer 43

• * make electricity
• Can be made from 2 different metals dipped in sale solutions of their own ions & connected by a wire (external circuit)

​

• How it works
  
  1. Zinc loses electrons more easily than copper. The zinc is oxidised to form Zu²⁺_(aq) ions thus releaseing electrons into the external circuit
  2. In the other half, the same # of electrons are taken from the external circuit, reducing the Cu²⁺ ions to copper atoms

Question 44

Good Reducing Agents

(Common)

Answer 44

• Active metals (sodium, magnesium, aluminium, & zinc)
  
  • relatively small ionization energies & low electro- negativities
  • Metal hydrides (NaH, CaH₂, & LiAlH₄) which formally contain the H^- ion are good too
• Hydrogen gas = reducing & oxidation agent
  
  • reduces non-metals
  • oxidises metals

Question 45

Common Oxidizing Agents

(Good)

Answer 45

• Flourine (F₂)
  
  • the strongest oxidizing agent
  • even water will burst into flame in its presence
• O₂, O₃, & Cl₂
  
  • good oxidising agents
  • 2^nd & 3^rd most electro negative elements
• High oxidizing states = good oxidizing agents
  
  • Permanganate (MnO₄^-)
  • Chromate (CrO₄^2-)
  • Nitric acid (HNO₃)
  • Perchloric acid (HClO₄)
  • Sulfuric acid (H₂SO₄)

Question 46

Ionic half-equations

Answer 46

• Ionic half-equations show oxidation or reduction
• Can combine them for different oxidizing or reducing agents together to make full equations for redox reactions
  
  • • ex. Magnesium burns in oxygen to form magnesium oxide
      
      • Oxygen is reduced to O^2-: O₂ + 4e^- ⇒ 2O^2-
      • Magnesium is oxidised to Mg²⁺: Mg ⇒ Mg²⁺ + 2e^-
        
        • (Both equations need to contain the same number of electrons so double everything)
      • 2Mg ⇒ 2Mg²⁺ + 4e^-
        
        • The electrons aren’t included in the full equation. There are 4 on each side so they cancel
      • 2Mg + O₂ + 4e^-- ⇒2MgO + 4e^-
      • Final = 2Mg + O₂ ⇒ 2MgO

Question 47

Redox Reactions: Oxidation States

(Rules)

Answer 47

• Oxidation states = Oxidation number

1. All atoms are treated as ions for this, even if they’re covalently bonded
2. Uncombined elements have an oxidation state of 0 (zero)
3. Elements just bonded to identical atoms (O₂ & H₂) also have an oxidation state of 0 (zero)
4. Ox. state of a simple monoatomic ion (eg Na⁺) is the same as its charge (eg 1)
5. In compounds or compoind ions, the overall ox. state is just the ion charge
   
   • ex. SO₄^2- - overall ox. state = -2
     
     • Ox state for O = -2 ( -2 x 4 = -8)
     • Ox state for S = +6 ( -2 = X - 8)
6. The sum of the oxidation states for a neutral compound is 0 (zero)
   
   • Fe₂O₃ - overall ox. state = 0
     
     1. ox st. for O = -2 (-2 X 3 = -6)
     2. Ox st. for Fe = -3 (-3 X 2 = -6
7. Combined Oxygen is nearly always -2
   
   • except
     
     • peroxides where it’s -1
     • Flourides - OF₂ = +2
     • O₂F₂ = +1
     • O₂ = 0
   • (In H₂O ox state of O = -2, H₂O₂ ox state = -1 b/c Hydrogen can’t give 2 electrons, only 1)
8. Combined Hydrogen is +1 except in metal hydrides where it is -1 ( and H₂ where it is 0
   
   • In HF ox st of H = +1
   • In NaH ox st of H = -1
9. Roman Numerals give oxidation state (ox #)
   
   • ex. Copper (II) sulfate ox state Cu = 2+
   • Manganate (VII) ion (MnO₄^-) ox state Mn = 7+

Question 48

Redox Reactions

Answer 48

• if electrons are transferred its a Redox reaction
• If electrons are lost its called Oxidation
• If electrons are gained its called a Reduction
• An Oxidising agent accepts electrons & gets reduced
• A Reducing agent donates electrons & gets oxidised
  
  • Na + ¹/₂Cl₂ ⇒ Na⁺Cl^-
    
    • Na is oxidised (loses an electron thus + in result)
    • Cl is reduced (gains an electron thus - in result)